91Ó°ÊÓ

The mass flow rate (\textdot{m}) can be calculated using the following formula: \[ \textdot{m} = \rho \times V \times A onumber \] Where: \rho is the density of the oxygen gas at the inlet (to be determined). V is the velocity at the inlet. A is the inlet area. First, use the ideal gas law to find the density \rho at the inlet: \[\rho = \frac{P}{RT} \] Using the given data: \( P = 30 \text{ lbf/in}^2 \) \( T = 440 \text{ R}\) \( R = 48 \text{ ft} \text{ lb}_\text{f} /(\text{lb}_\text{m}\text{ R}) \) Now, convert the pressure to \( \text{ lb}_\text{f} / \text{ ft}^2 \): \[ P = 30 \text{ lbf/in}^2 \times 144 = 4320 \text{ lbf/ft}^2 \] Now find the density: \[ \rho = \frac{P}{RT} = \frac{4320 }{48 \times 440} = 0.2045 \text{ lb}_\text{m}/\text{ft}^3\] Now calculate the inlet area in square feet: \[ A_\text{in} = 2 \text{ in}^2 \times \left( \frac{1 \text{ ft}^2}{144 \text{ in}^2} \right) = 0.0139 \text { ft}^2 \] Now, find the mass flow rate: \[ \textdot{m} = 0.2045 \text{ lb}_\text{m}/\text{ft}^3 \times 950 \text{ ft/s} \times 0.0139 \text{ ft}^2 \approx 2.705 \text{ lbm/s}\]

According to the continuity equation for steady flow, mass flow rate is conserved: \[ \textdot{m}_{in} = \textdot{m}_{out} \] So, \[ \rho_{in} V_{in} A_{in} = \rho_{out} V_{out} A_{out} \] Use the area ratio to find the exit area: \[ A_{out} = 15 \times A_{in} = 15 \times 0.0139 \text{ ft}^2 = 0.2085 \text{ ft}^2 \] Rearranging the above equation: \[ \rho_{in} V_{in} A_{in} = \rho_{out} V_{out} A_{out} \] Solving for \( \rho_{out} \): \[ \rho_{out} = \frac{\rho_{in} V_{in} A_{in}}{V_{out} A_{out}} = \frac{0.2045 \times 950 \times 0.0139}{25 \times 0.2085} \approx 0.515 \text{ lbm/ft}^3 \]

Use the ideal gas law again to find the exit temperature: \[ \rho = \frac{P}{RT} \] Solving for \( T_{out} \): \[ T_{out} = \frac{P_{out}}{\rho_{out} R} \] But first, apply the steady-flow energy equation: \[ h_{in} + \frac{V_{in}^2}{2} = h_{out} + \frac{V_{out}^2}{2} \] For ideal gases: \[ h = c_p T \] Assuming constant specific heats: \[ c_p = 0.240 \frac{Btu}{lbm R} \approx 1.71 \text{ Btu/lbm R} onumber \] So, \[ (1.71 \times 440 + \frac{950^2}{2 \times 32.174}) = 1.71 T_{out} + \frac{25^2}{2 \times 32.174} \] Solving for \(T_{out}\): \[ 1.71 T_{out} = 924 + 14046.3 - 9.7 \] \[1.71 T_{out} = 14962.6 \] \[ T_{out} \approx 437 \text{ R} \]

Use the ideal gas law again: \[ P_{out} = \rho_{out} R T_{out} \] Substitute the values: \[ P_{out} = 0.515 \times 48 \times 437 = 1077.9 \text {lbf/ft}^2 \] Now, convert back to lbf/in^2: \[ P_{out} = \frac{1077.9}{144} = 7.48\text {lbf/in}^2 \]